In a series LCR circuit, \( C = 80\,\mu\mathrm{F} \), \( L = 5.0\,\mathrm{H} \) and \( R = 40\,\Omega \) are connected to a variable frequency source of 240 V. The angular frequency of the source which drives the circuit at resonance will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 50\,\mathrm{rad\,s^{-1}} \)
In a series LCR circuit resonance occurs when \( X_L = X_C \), that is \( \omega L = 1/\omega C \). The resistance and the source voltage do not enter.
\( \displaystyle \omega_0 = \frac{1}{\sqrt{LC}} \)
\( \displaystyle C = 80\,\mu\mathrm{F} = 80\times10^{-6}\,\mathrm{F} \)
\( \displaystyle \omega_0 = \frac{1}{\sqrt{5\times80\times10^{-6}}} = \frac{1}{\sqrt{4\times10^{-4}}} = \frac{1}{2\times10^{-2}} = 50\,\mathrm{rad\,s^{-1}} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – “… 7.8 Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω. (a) Determine the source frequency which drives the circuit in resonance.”